Using Cofactors of elements of third column, evaluate.
To evaluate a determinant using cofactors, Let
B =
Expanding along Row 1
B =
B = a11 A11 + a12 A12 + a13 A13
[Where Aij represents cofactors of aij of determinant B.]
B = Sum of product of elements of R1 with their corresponding cofactors
Similarly, the determinant can be solved by expanding along column
So, B = sum of product of elements of any row or column with their corresponding cofactors
Cofactors of third column
A13 = (-1)1+3 × M13 = 1 × = 1 × (1 × z – 1 × y) = (z – y)
A23 = (-1)2+3 × M23 = (-1) × = (-1) × (1 × z – 1 × x) = - (z - x) = (x - z)
A33 = (-1)3+3 × M33 = 1 × = 1 × (1 × y – 1 × x) = (y – x)
[Where Aij = (-1)i+j × Mij, Mij = Minor of ith row & jth column]
Therefore,
Δ = a13A13 + a23A23 + a33A33
Δ = yz (z - y) + zx (x - z) + xy (y - x) = z [y (z - y) + x (x - z)] + xy (y - x)
Δ = z (yz - y2 + x2 - xz) + xy (y - x) = z [(yz - xz) + (x2 - y2)] + xy (y - x)
Δ = z [z × (y - x) + (x + y) × (x - y)] + xy (y - x)
Δ = z × (y - x) × (z – x - y) + xy (y - x)
Δ = (y - x) × (z2 – xz – yz + xy)
Δ = (y - x) × [z (z - x) – y (z - x)] = (y - x) × (z - y) × (z - x)
Δ = (x - y) (y - z) (z - x)
Ans: Δ = (x - y) (y - z) (z - x)